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(C) The coordinates of the centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are given by the formula:$G = \left(

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$1$

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(C) The coordinates of the centroid $(G)$ of a triangle with vertices $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are given by the formula:$G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$Given vertices are $(a, 2)$,$(2, -a)$,and $(0, 2)$.Substituting these values into the formula:$G = \left( \frac{a + 2 + 0}{3}, \frac{2 - a + 2}{3} \right) = \left( \frac{a + 2}{3}, \frac{4 - a}{3} \right)$According to the problem,both coordinates are equal:$\frac{a + 2}{3} = \frac{4 - a}{3}$$a + 2 = 4 - a$$2a = 2$$a = 1$

If both the coordinates of the centroid of a triangle with vertices $(a, 2)$,$(2, -a)$,and $(0, 2)$ are equal,then $a=$...........

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